Let’s say that our converter is a half-bridge converter.

The input voltage to the converter will vary from 150VAC (212VDC) to 250VAC (354VDC). Output voltage of the converter is 14VDC. The switching frequency is 50kHz.

Transformer primary: 26 turns

Transformer secondary: 4+4 turns

The formula for calculating the minimum required inductance is:

We need to calculate the output voltage at the transformer secondary at 354VDC input, which is our maximum input voltage.

We’ll assume that the voltage drop due to the rectifier diode is 1V. Thus the average output voltage at the transformer secondary is 15V. Transformer turns ratio (primary : secondary) = 26:4 = 6.5

So, when average secondary voltage equals 15V, average voltage across transformer primary is 6.5 * 15V = 97.5V. If duty cycle was 100%, voltage across transformer primary would be 177V (half the DC bus voltage – think half-bridge topology). So, the duty cycle is (97.5/177)*100% = 55%.

Average output voltage at transformer secondary is 15V with a duty cycle of 55%. Thus, peak output voltage is 15V/0.55 = 27.3V, after which we have assumed a diode drop of 1V. So, Vin(max) is 26.3V.

At maximum input voltage, duty cycle will be lowest. This is when the off time will be the highest.

We have calculated a duty cycle value of 55% – this is the lowest duty cycle value. As switching frequency is 50kHz, time period is 20µs. The off time is 0.45 * 20µs = 9µs. That is our Toff(est).

Let’s say that the minimum load will draw 500mA current. With a 14V output and 500mA current, the power dissipated in the output resistor will be:

P = VI = 14 * 0.5 W = 7W

That is a lot of power! If it’s acceptable, go ahead and use a 500mA minimum load. If you choose to bring the minimum load down to 250mA, you bring down power dissipation (above) to 3.5W.

So we now have all required parameters. Let’s plug them into the formula.

This is the minimum required inductance. You should use an inductance larger than the minimum value calculated, since, well you calculated the *minimum required* inductance.

Let’s say we’ll use an inductance of 450µH. Let’s say that we’ve selected a toroid core with an AL value of 64nH per turn squared.

Firstly, the required inductance is 316µH which is equal to 316000nH.

Thus the required number of turns is:

You can use either 70 or 71 turns. This is for 316µH.<

For 450µH:

Round this up to 84 turns.

There we have it. You can use this simple method to calculate the required output inductance for a converter that uses the forward, push-pull, half-bridge or full-bridge topology. It’s simple and I hope I’ve been able to make you understand clearly.

Original article by Tahmid at his blog.